Replacement 1n3996 in AA Xovers?

[nk2j]

Hi i'm working on upgrading the AA crossovers in my lascalas(Ser # 4N782) and wondered if replacing the two diodes might not also be beneficial?... and what a good current tech replacement might be?

also since i will ne getting a pr of cornwalls this weekend.. (the Cornwall E's that were just on eBay) i wondered about the difference between the cornwall and lascala crossovers...i would like to find a schematic of the cornwall xover (serial# 14K301)

tks,

john

Equipment:

various DIY SET amps; currently: SRPP 6sn7/27/45, 6j5/10y, DC Darling, etc.

ole Dynaco PAS-3 upgraded preamp

Lascalas,... Cornwalls coming..(can't wait! lol)

DIY pure silver speaker cables, Radio shack Fusion interconnects

Cal audio transport, Timbre TT-1 DAC

[BEC]

I don't believe there is any reason to expect the zeners are bad. I have managed to find an exact replacement if you do want to replace them. Replacing the old caps is, however, a very good thing to do.

Bob Crites

[nk2j]

thanks Bob. i expect you are right about the zeners... i suppose zeners are pretty much zeners lol... i haven't found any 13 uf caps yet though.. and although i'm sure the inductors klipsch used are ok the bigger wire gauge ones currently available would likely be better..

[JohnA]

I would NOT replace or use the zener diodes in the Type AA. They sound awful when they short out the power amp trying to protect the tweeter. I'd substitute a polyswitch and 100 ohm or greater resistor (in parallel and the pair in series with the tweeter). You can buy them from Parts Express. This is what Klipsch uses in later crossovers.

[BEC]

John,

Why do you say that the zeners would short out the amp? If I am thinking correctly, when the input power to the crossover gets to around 100 watts, and the tweeters are being sent two watts (2 percent of the power) the zeners would begin to conduct. When the zeners conduct at 2 watts (200 watts into the crossover) the impedance of the zeners would be equal to the impedance of the tweeter and the combined impedance would be 4 ohms. Still not a short. They are not like a switch, more like a variable resistor.

Bob Crites

[Deang]

nk2j, Tom Mobley on 2-channel says these are 1972 models, with K-33, K-55-V, alnico K-77. This means you have the Type B crossover. Just the autoformer and woofer inductor -- with a 4uF and 2uF cap.

Incidently, Bob has some nice (and affordable) polypropylene in oil motor runs, and would probably be willing to sell you the needed values for your projects. I can supply the same values in Jensen PIOs if you like to spend money and dabble in the black arts.

[nk2j]

Thanks for the input guys.. its much appreciated... i am glad to know the date (1972) of them as i really had no idea..

also thanks for the info on the crossover caps.. i'll be doing my Lascalas too so i ordered the caps from Bob for both cornwalls and lascalas..

might probably try the jensens at a later date after a few more pension checks lol

[JohnA]

Hey Bob,

It's been a while since I looked at the curves on a zener diode. As I remember, once the breakdown voltage is exceeded the current flow increases dramatically, meaning the resistance goes way down. While these diodes might exhibit 4 ohms each above breakdown, I doubt they do. The result is a short to all power above 2 watts across the tweeter. If the 2 diodes only dropped to 4 ohms each, they would absorb 1/2 of the applied power and could not limit the tweeter's power any further.

Besides, they sound nasty as $%^& when they operate!

[Al Klappenberger]

John,

I think your's right. Zeners hold a constant voltage across them no matter the current. Ohms law says therefore that their resistance varies according to the current through them. They are open circuits when the voltage is below the brakedown pint and do nothing. I don't believe that they can be considered 4 ohms at any but some instantanious point in time. They should be considered a short circuit when they come into action.

Al k.

[BEC]

John and Al,

Think about what we are saying about how the zeners act. If they maintain a constant voltage, by logic, they are variable in their conduction. If voltage tries to increase further, they conduct more heavily to maintain the constant voltage. Also remember that this conduction starts at the voltage equivalent to 100 watts of crossover input. Tough to imagine a change in the tweeter output being something you could hear while everything in the speaker is being hammered by 100 watts. I think any changes will be drowned out by the shaking around of the furniture in the room and such.

Bob

[Shock-Late]

In any case, the network sounds better with the zeners diconnected... it's quite sublte, but you can hear it.

[Al Klappenberger]

Bob,

Your assumption that the input to the network is going to be 100W before the

diodes conduct is off the mark a bit. The actual input level when the diodes

conducts is a function of the frequency and the slope of the tweeter filter

when there is no distortion. If the offending signal is at or above about 6

KHz it would only take a few watts to drive them into the breakdown region.

What you are actually saying is that a tweeter protector is not necessary at

all if you just keep your lips off the volume control! The big factor is

actually clipping by overdiving a medium powered amp and the compression that

happens to the lows when you do. The lows square off generating harmonics in

the tweeters range while the highs that belong there, even though they are

below the clipping level, get larger because the volume is jacked up. The

total input from the amp may only be 20 or 30W but the tweeter sees a huge

proportion of it. The result is a dead tweeter or conducting zeners. If you

are using an amp that will deliver 100 watts of CLEAN power, your statement

would be true though.

Al K.

[BEC]

Al,

The zeners in the AA crossover limit the amount of power to the tweeter in the 100 watt continuous rated K-horn to 2 watts. You can of course change the signal easily to no longer fit into this 2 percent relationship. Most amps have a treble boost that can easily add ten or more db to the HF signal. In that case (with a lot of treble boost) a 20 watt amp may be too much for the tweeter.

In fact though, it matters not at all what frequency the signal is that gets to the tweeter as far as the zener protection scheme is concerned. If two watts get to the tweeter. the zeners that are in parallel with the tweeter start to conduct. Lets hope that what ever power amp we use is "clean", but the zeners do not care. They will conduct clipped signals to protect the tweeters as well as nice linear ones.

By the way, the impedance of the pair of back to back zeners is around 4.5 Kohms at 10 khz. It is hard for me to get my mind around the idea that some one can hear the difference between them being in or out unless they are being driven into conduction. I am not saying that someone can't "hear" them. I am saying that I can't "hear" them and that I can't see the physical mechanism that should allow me to "hear" them unless they are driven into conduction.

While we are on this one, let me ask you a question. Why would you consider a "clipped" signal of a given level more detrimental to a tweeter than a clean signal of that same level? I have trouble seeing the mechanism for one being more damaging to the tweeter than the other.

Bob

[Al Klappenberger]

Bob,

It's pretty difficult to explain why a distorted signal is destructive

because it is what you might call a comedy of errors that does the damage. I

had hoped to explain it in my last post. I guess I didn't do a very good job!

Lets start with your 100 W input assumption. If I understand you, you have

100W of power at, say 100 Hz. and maybe 2W of power at 7 Khz. Clearly the

tweeter sees all of the 7 Khz since the tweeter filter has very little loss.

The woofer is taking the 100W at 100 Hz. That's fine. There should be no

smoke or fire anywhere if it's only sustained for a shor time. Now:

Somebody's exuberant son turns up the volume control another 3 dB. Now the

amp is called upon to deliver 200W at 100 Hz and 4 W at 7 KHz. 4 watts alone

will do enough damage itself, but the amp is now driven to clipping on the

100 Hz signal. At that point each sine wave component is squared off

generating power at harmonic frequencies all the way up to and beyond the

tweeters range adding to the 4 clean watts that should be there! Poof! no

more tweeter! If your amp is rated 20W you are far more likely to run it too

hard causing the clipping than if it was rated at 100W. The 20W amp will

still be able to deliver the 4 clean watts at 7 Khz, BUT the distortion

products caused by the clipping of the 100 Hz signal are added and also seen

by the tweeter. This is called "compression". The quick answer is that a

clipped sine wave becomes a square wave of far more total energy than the

sine wave had even if the peak-to-peak voltage is the same. The peak-to-peak

voltage of the fundamental IS actuall still virtually the same! The extra

power is in the poor tweeters frequency range! A sine wave has RMS voltage =

.707 times peak. It's the RMS value that is the heating component of a

waveform. A square wave has RMS equal to peak. Increasing the level at

clipping does not increase the peak-to-peak level of the signal, it just

raises the RMS value and therefore the total power. That added power is in

high frequency harmonics that toast the tweeter. I think you have a spectrun

amalyzer. Try hooking it to an amplifier output and overdrive it. You will

see the fundamental stay nearly the same while the harmonics jump WAY up.

BTW: I think the audible portion of the zeners conducting is when extreme low

frequency signals come through the tweeter filter at a high enough level to

cause the harmonics they generate when the diodes conduct to be within the

range of hearing. I don't think you can hear clipping of a 6 KHz signal. The

harmonics would be mainly 18 Khz and higher. This would happen big time if a

zener protector was used with the "A" network that only has at single cap for

a tweeter filter. It could also damage the amp. I think this is why the

zeners gave way to the poly-switch protection method in later networks.

Al K.

[BEC]

Al,

Remember that the zeners are there. They are the subject of this discussion. How did you get your 4 watts past them? If you assume that signals are all clipped to the point the that peak is the same as effective, that still isn't 4 watts. By the way, all K-77 diaphragms made since the change from aluminum VC windings to copper in the early 70's have been rated for 5 watts continuous.

[Al Klappenberger]

Bob,

I thought you asked why clipping would cause damage to a tweeter. I answered with the idea that the zeners were not there! Maybe I should have assumed you already understood that and were asking why the zeners wouldn't save the tweeter under amp clipping coditions. Yes, they will. The zeners do a fine job of protecting the tweeter. The problem with them is that they short out the amp when they operate! Some amps can't handle that. They can also cause audable distortion if extreme high power lows get through the tweeter filter at a high enough level to cause them to conduct. That may be a good thing if it reminds you that you have the volume up to high and that it's time to cool it! It's certainly better than a cooked tweeter. All the numbers I picked were off the top of my head too. I didn't even think about the 2W or 5W tweeter power ratings. Tweeter survival all depends on the tweeter filter, the particular tweeter, the diode breakdown voltage and the like.

Al K.

[BEC]

Al,

We seem to have come full circle back to the "zeners short the amp" concept. How can a pair of back to back 5.1 volt zeners which are designed to maintain a peak to peak voltage of 10.2 volts across them be considered a "short" of any kind?

Bob

[JohnA]

Bob,

First, diodes are not constant voltage devices, at least to my way of thinking. The zeners Klipsch used do not maintain a voltage. I like to view diodes as check valves. Current can flow one direction, but not the other. A zener diode is like a check valve with pressure relief bypass. Current flows freely one way and not the other, unless the voltage exceeds breakdown. When it does, the pressure relief allows essentially free flow in the other direction. It the current is sent to ground through another diode turned the opposite direction, it sees essentially no resistance going to ground, thus a short.

Zener diodes are voltage limiters, not constant voltage regulators.

In the AA network, when the voltage is below 5 volts, one of the zeners lets the current from the positive half of the wave pass to the other diode where it is blocked. The pair do the same to the negative half of the wave.

When the voltage exceeds 5 volts on either half of the wave, one diode passes just as it always did, but the other, because the wave is 5V or higher, also passes the current. That makes a dead short.

There is no specific correlation to system input power. I'm sure that is adding confusion. Even if 100 watts is applied to the system in a song like Edgar Winter's "Frankenstien", waay more than 2 watts would be applied to the tweeters because of the Highs contained in the synthesizer riffs. Your 2 watt/100 watt relationship would seem logical for classical music and many others played on natural instruments.

The analysis needs to be based on the power applied to the tweeter circuit no matter how it got there.

[Al Klappenberger]

Bob,

John explained the short concept pretty well. In truth a short circuit can be

virtually any value. It would depend on the circuit in which the "short" is

applied. To and automobile battery, for example, a short circuit would be a

resistance equivalent to that of a jumper cable connected right across the

terminals. On the other hand, a resistance of a few thousand ohms would be

all it would take to "short out" a high impedance circuit like the grid of a

vacuum tube or the high voltage second anode circuit of a TV set picture

tube. A zener diode has a changing resistance depending on the amount of

current that is flowing through it. A simple series regulator consisting of a

series resistor from a voltage source and zener diode connected to ground

uses this property to operate. The zener holds a constant voltage and the

series resistor limits the current. Current is then shunted from the zener to

the load connected across it. Without the resistor the diode would blow up

with a bang the instant it was subjected to a stiff voltage source greater

than its breakdown voltage. If you look at the schematic of the AA network

you can easily find the zeners, but where is the series resistor that limits

the current? There answer is that it is IN THE AMP! Then it becomes a

question of which will give out first, the zener or the amp! As I said

before, it depends on the amp. Some can take it and some can't.

BTW: The P-P voltage across the two 5.1V zeners is NOT 10.2V, it's about 5.7V.

One of the diodes is operating in the forward direction which maintains about .6

to .8V while the other is breaking down in the opposite direction

contributing 5.1V. The sum is 5.1 + .6 = 5.8 (roughly).

Al K.

[BEC]

"First, diodes are not constant voltage devices, at least to my way of thinking.  The zeners Klipsch used do not maintain a voltage.  I like to view diodes as check valves.  Current can flow one direction, but not the other.  A zener diode is like a check valve with pressure relief bypass."

John,

The most often found use in my experience for a power zener (like the 1N3996) is as a voltage regulator. When operated inside the area of its reverse conduction it maintains a constant voltage from its "knee" to its maximum current rating. In the case of the 1N3996 we are discussing, that maximum is 1780 mA continuous or 10 watt dissipation. Small signal zeners would never be operated this way as they have almost no ability to dissipate heat from continuous operation. The small ones would more fit your description of being used as "check valve with pressure relief bypass".

Don't beat me up too much on the "2 percent of the power to the tweeter" subject. PWK said it was more like one percent, but he designed the AA crossover to allow 2 percent of the Khorn's rated power to get to the tweeter. A bit to late to take that subject up with him though. He did, however, leave a lot of writing to justify his position on this and about everything else he did in his designs.

Bob